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3 years ago

Section 1: Experimental Overview

Physics

1 answer:

Luden [163]3 years ago

7 0

Answer:

1-state what the lab is about, that is, what scientific concept (theory, principle, procedure, etc.) you are supposed to be learning about by doing the lab. You should do this briefly, in a sentence or two. If you are having trouble writing the opening sentence of the report, you can try something like: "This laboratory experiment focuses on X…"; "This lab is designed to help students learn about, observe, or investigate, X…." Or begin with a definition of the scientific concept: "X is a theory that…."

2-give the necessary background for the scientific concept by telling what you know about it (the main references you can use are the lab manual, the textbook, lecture notes, and other sources recommended by the lab manual or lab instructor; in more advanced labs you may also be expected to cite the findings of previous scientific studies related to the lab). In relatively simple labs you can do this in a paragraph following the initial statement of the learning context. But in more complex labs, the background may require more paragraphs.

Explanation:

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What is someones target heart rate at 70% at age 17

baherus [9]

Answer: normal

Explanation:

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3 years ago

An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances o

Serhud [2]

Answer:

$Q_t= 8.3 * 10^3 C$

Explanation:

From the question we are told that:

Resistor $R=1000ohms$

Voltage $v=120_V$

Capacitance of c_1 $c_1=20 \mu F$

Capacitance of c_2 $c_2=60 \mu F$

Time $t=0$

Generally the equation for charges is mathematically given by

$For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\$

$ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3$

Generally the equation for voltage across capacitors is mathematically given by

$V_c(t)=V(1-e^{-t/RC})$

$C=C_1+C_2=80 \mu f\\t=2RC=>160000s$

$V_c(t)=120(1-e^{-(160000)/1000*(80)})$

$V_c(t)=103.7598$

Generally the equation for charges is mathematically given by

$Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\$

Generally the equation for total charges $Q_t$ is mathematically given by

$Q_t=Q1(t)+Q2(t)$

$Q_t= 8.3 * 10^3 C$

5 0

3 years ago

Use the information below to answer questions

Ulleksa [173]

Answer:

The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁ = 5 × 10⁻⁸ C - q₂ (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

$q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\ = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16} - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8} - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C$

q₁ = 5 × 10⁻⁸ C - q₂

q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C

5 0

4 years ago

If your front lawn is 16.016.0 feet wide and 20.020.0 feet long, and each square foot of lawn accumulates 13501350 new snowflake

asambeis [7]

Answer:

The mass of snow accumulated in the lawn in 1 hour equals 57.024 kilograms

Explanation:

Given

Width of lawn = 16 feet

Length of lawn = 20 feet

Thus total area of lawn equals $16\times 20=320sq.feet$ \

Now it is given that 1 square foot accumulates 1350 new snowflakes each minute thus number of snowflakes accumulated by 320 square feet in 1 minute equals

$320\times 1350=432000$

Now it is given that average mass of each snowflake is $2.20mg=2.20\times 10^{-3}g=2.20\times 10^{-6}kg$

Hence the mass accumulated per minute equals $432000\times 2.20\times 10^{-6}=0.9504kg$

Now since there are 60 minutes in 1 hour thus the mass accumulated in 1 hour equals $0.9504kg\times 60=57.024kg$

4 0

3 years ago

The rms current output in a circuit is 7.75 A. What is the maximum current?

Andru [333]

The maximum current in an ac circuit is related to the rms current by:
$I_{rms} = \frac{I_0}{ \sqrt{2} }$
where $I_{rms}$ is the rms current and $I_0$ is the maximum current.

If we re-arrange the formula and we use $I_{rms}=7.75 A$ , we can find the value of the maximum current:
$I_0 = \sqrt{2} I_{rms} = \sqrt{2}(7.75 A)=11.0 A$

5 0

3 years ago